Max™ is firmly convince that with his understanding of relativity, He really could, by traveling really fast, leave Earth, and get back before he left. Now I have nothing against him believing this fantasy.

I just want an answer to this scenario: Suppose I were to leave Earth and travel to Mars, instantly. Then suppose I spend an hour having coffee with his girl friend, Miss Bright, who is in confinement there. I then return Instantly to Earth. How long am I gone from Earth? What speed is there whence I could get back to Earth, before I left?

Suppose I traveled at the speed of twice instant. Would that work?

Bob S and CY -

While Xam could see you, by virtue of viewing the light horizon of you drinking from your cup of coffee...Xam can never go fast enough to be there on Mars when you are sipping, once you slurp down your last drop.

I suspect Xam stills lumps the event ( drinking of the coffee )
with the after event horizon of the now completed event.

...........................................................

"wikipedia" btw

"The Galilean transformations assume that time is the same for all reference frames."

did people miss this too from wiki...

"For example, in 1D, if the mapping is a translation of 3 to the right, the first moves the origin from 0 to 3, so that the coordinate of each point becomes 3 less, while the second moves the origin from 0 to -3, so that the coordinate of each point becomes 3 more."

...............................................................

Does anyone else
( except max...I am not longer gonna respond to max's logic/comments )
see a vt here ?

Do we all get that if a points' abscissa = 3.475869 in one frame it HAS to have an abscissal value of 3.475869 in the other frame. That is the very definition of what a coordinate transformation is. Really dudes, this is not rocket science here.

Agreed Bob S?

Does this not mean that the point X' of coordinate length 4 in s',
is not 4 in S ?

Although 11-7 does = a line segment in S of length 4,
it is no longer representative of what an absccisa/coordinate is.

Point P' in s is without abscicca. It is not a coordinate therefore.

Point P', belongs at 4,0 in s....this then would be a viable coordinate transformation. Einstein's is half-assed as he leaves S' with a coordinate
and fails to transform it and maintain its coordinate status...into S.
Read the wiki thing above a few times. Do you notice that when X' is placed at 4,0 in S that x = x' + vt still works ? Isn't that weird. (rhetorical)

Because, this time, with P' at 4,0 in S...all three ARE coordinate lengths
whereas P' at 11,0 in S, only has two of three as coordinate lengths in their OWN frame.

steve waterman

Well technically no one can move faster through spacetime than light, so the point is, as they say, moot.

Also: lol, Steve, sneaking in that argument again.


Point P' in S' doesn't have two absiccal values in S', or S.

As I said.


From Galileo's Bed the Telescope is 11 feet away.

From Galileo's Desk the Telescope is 4 feet away.


The Telescope is not 4 and 11 feet from the Desk.

The Telescope is not 11 and 4 feet from the Bed.

Egg? Grease? Shaving cream? Moot juice?

Do you realize that speed cannot get you back before you left? It's not moot.

Do you realize the future cannot exist until the immediate present has come into being?

Do you realize the past wave fronts of light from an event cannot ever again be seen by those at the event, baring instantaneous travel?

You should try Mr. Waterman's Universe. At the least, his statements are not illogical. http://watermanpolyhedron.com/

Or else, go there and find all the illogical stuff, and let me know, so I can understand what a fool I am for thinking it is logical.

He claims that the Telescope, when viewed from the Desk, is 4 feet away.

He then claims that now, from the Bed, the Telescope is both 11 feet and 4 feet away.

Technically there is a sense in which this is correct, in that 4 feet is a component of 11 feet.

The problem being, you need to give a reference point to explain why you are arguing that from the Bed, the Telescope is now 11,0 and 4,0.


Without invoking any possibly obscure mathematical terms, that is the crux of the problem.


Steve says that Point P', the Telescope viewed from the Desk, now has two values (11,0) and (4,0) in the reference frame S, the Bed.


I contend that it only has one value from the Bed, or the composite value including the motion from the Bed to the Desk (x' = x - vt) but not both at the same time.


I contend that the Telescope from the Bed is P. (11,0)


All the other arguments about lines and points are missing this simple objection he had.

Steve,
"Do we all get that if a points' abscissa = 3.475869 in one frame it HAS to have an abscissal value of 3.475869 in the other frame. That is the very definition of what a coordinate transformation is. Really dudes, this is not rocket science here.

Agreed Bob S?"

I agree Steve, if not equal, then it is not an inertial reference frame to be compared.

bob s

bob s -

Thank you for your feedback.

Talk to me Bob. Since you get what a transformatiom does...,
do you also see that allowing x = x' + vt is wrong, just as
x = x' + d would be wrong....
as P' in S should be at 4,0 not 11,0?

Do you see that the coordinate length of x' of 4 in S', becomes
11 not 4, by placing P' at 11,0 in S ?

........................................................
Given, of course that...

frame S
has O at 0,0
P at 11,0

frame S'
has O' at 0,0
has P' at 4,0

after P' >>>>>>> frame S

then...

frame S
P at 11,0
P' at 11,0
P = P'

However, Einstein's x = x' - vt, means here, with the above notation,

P' = P - S' or 4 = 11 - 7.
Although this obviously looks right...since 4 + 7 = 11....
it is part of why this theory has snuck by for oh so many years.

Since
P' = P - S' AND P' = P

therefore,

one point [ P at 11,0 ]
now has two abscissa [ P at 11,0 and P' at (11,0 - 7,0) ]
in one frame [ S ]

Do you agree with this above statement Bob ?

as the length of 4 has now also attributed to P' in S.

steve waterman

one point [ P at 11,0 ]
now has two abscissa [ P at 11,0 and P' at (11,0 - 7,0) ]
in one frame [ S ]

It doesn't have two abscissa in one frame.

That is the sum of all this confusion.


In frame S, P = 11,0.

P' = P
S' =/= S


Going to my above, Galileo appropriate, description.


The Bed is frame S.

The Desk is frame S'.


From the Bed, the Telescope is 11 feet away. This describes P.

From the Desk, the Telescope is 4 feet away. This describes P'.

In the frame of the Bed, the Telescope is not 11 feet, and 11 feet - 7 feet away.

It is just 11 feet away.

In the frame of the Desk, the Telescope is 11 feet - 7 feet away, not 11 feet away.

Steve,
"therefore,
one point [ P at 11,0 ]
now has two abscissa [ P at 11,0 and P' at (11,0 - 7,0) ]
in one frame [ S ]

Do you agree with this above statement Bob ?
as the length of 4 has now also attributed to P' in S.

steve waterman"

I agree in principal Steve, but not in application. Reason, if you overlay S with S' you would have two abscissa "[ P at 11,0 and P' at (11,0 - 7,0) ]" but it would require a new frame reference such as S''.

bob s

Bob S -

"I agree in principal Steve, but not in application. Reason, if you overlay S with S' you would have two abscissa "[ P at 11,0 and P' at (11,0 - 7,0) ]" but it would require a new frame reference such as S''."

No, not at all. Instead of adding a third reference frame S", just forget about S' ( and S") completely... after point P' is transformed into S.

So, just frame S, one frame, after the transformation the transformed point P' would have two abscissa in S, in principle and application.

To even be a coordinate transformation,
P' must be at 4,0 in S just as it is in S'.
According to the very definition of what a coordinate transformation does.

No point was ever transformed/changed/moved....as P = P' in both frames.

Only Einstein counts backwards from the point to [ ooops...not to the origin, my error...towards the origin ], instead of from the origin to the point... in the entire history of Mathematics!
11 - 7 = 4 It alone, should make one think...and raise an eyebrow or two.

You see, being mathematical, there is no proviso involved...either the math stands or it falls...so, one frame then only...after the point be transformed.

steve waterman

Steve, "To even be a coordinate transformation,
P' must be at 4,0 in S just as it is in S'.
According to the very definition of what a coordinate transformation does."

Following the logics of maths Steve, your statement should be self evident.

As to my (S''), and I won't belabor the point, but if S and S' are over laid with each other then the result would be a third coordinate which, if not identified, might create confusion to a third observer, consider, a person observes S and then observes S' and is then shown a composite which is identified as S it would take an explanation as to why the original S and the new S do not appear the same therefor, giving the new coordinate a new designation, such as S'' would eliminate any confusion. No big deal, just my idea.

bob s

Bob S -

Steve, "To even be a coordinate transformation,
P' must be at 4,0 in S just as it is in S'.
According to the very definition of what a coordinate transformation does."

Following the logics of maths Steve, your statement should be self evident.

Steve So that is a yes then ? Please.... do i need to pull teeth here ? Should i just assume that you agree with every other point i made then in my previous entry, that received zip for a comment ?...not even a damn agree or disagree.

"As to my (S''), and I won't belabor the point, but

Steve Too late, this is not about observers or a third reference frame. Please do me the favor of letting me know what if anything you are not in full agreement with. Mainly, 1 frame has 1 point with two assigned and yet unequal abscissal values ? A simple yes or no then at least to what i contend in my previous entry...and except for one statement of mine...just blew off all the rest without an ounce of feedback....how about those agree or disagrees with all those other points ? I fear you are still swimming up the Nile...and not on board the x = x'+ vt is wrong boat - even a tad.

steve waterman

No point was ever transformed/changed/moved....as P = P' in both frames.


The origin of S was transformed to the origin of S'.

Sorry, to be more specific.

The coordinate descriptions from the origin of S were transformed to the descriptions from the origin of S'.


Claiming that S views P' is fundamentally wrong.

S views P.

P' does not exist in S as a coordinate set, unless you ask someone at S 'what is the absiccal value of P' in the S' coordinate set?'

Then they will say "That depends on how fast S' is and how far away it is, let's use the Galilean to determine that."


Then you say 'Uh, no, P' has two values in S, the Galilean is wrong.'

Then they will look at you funny.

Steve, I did agree with you, sorry if you mis-understood.

bob s

WHY!

Why would you agree with that?


The Telescope is not 4 and 11 feet from the Bed.

Think about what you're saying!

Bob S -

Steve, "To even be a coordinate transformation,
P' must be at 4,0 in S just as it is in S'.
According to the very definition of what a coordinate transformation does."
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
P' is 4,0 + vt in S, it doesn't exist in S by itself.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Following the logics of maths Steve, your statement should be self evident.

Steve So that is a yes then ? Please.... do i need to pull teeth here ? Should i just assume that you agree with every other point i made then in my previous entry, that received zip for a comment ?...not even a damn agree or disagree.

"As to my (S''), and I won't belabor the point, but

Steve Too late, this is not about observers or a third reference frame. Please do me the favor of letting me know what if anything you are not in full agreement with. Mainly, 1 frame has 1 point with two assigned and yet unequal abscissal values ? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Why would you claim that? 1 frame has 1 point with 1 assigned abscissa, and the other frame has 1 point with 1 assigned abscissa.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


A simple yes or no then at least to what i contend in my previous entry...and except for one statement of mine...just blew off all the rest without an ounce of feedback....how about those agree or disagrees with all those other points ? I fear you are still swimming up the Nile...and not on board the x = x'+ vt is wrong boat - even a tad.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

WRYYYY!!

KHAN! DO YOU HEAR ME? DO YOU!

KKKKKKKKHHHHHHHHHAAAAAAAAANNNNNNNNNNNNNNNN!!!!!!!!!!!!!

http://www.youtube.com/watch?v=LTMI6v9NyJg&feature=related

Max, I am not going to engage you in this string.

bob s

Me too.

steve waterman

WHY!

I've yet to see a good argument about that.


P' does not have two values in S.

P' in S is P which has one value.



Use the language I did.



Bed 1 2 3 4 5 6 7 8 9 10 Telescope

Bed -6 -5 -4 -3 -2 -1 Desk 1 2 3 Telescope

Bed 1 2 3 4 5 6 Desk 8 9 10 Telescope

How can you get that the Telescope is at 4,0 from the Bed?

Apparently, no one else is from your country.

max-

I likely will regret this, however, I am gonna try
to splain this again...I do believe that you are ignoring
the definition of a coordinate transformation and its example.

Third time posting this....( no agreement yet by you on this example )

"For example, in 1D, if the mapping is a translation of 3 to the right, the first moves the origin from 0 to 3, so that the coordinate of each point becomes 3 less, while the second moves the origin from 0 to -3, so that the coordinate of each point becomes 3 more."

your example...

Telescope is four from Desk in S'.
After transformation, the POINT telescope,
must be at four from the Bed in S.

Since this has not happened...then point Telescope was never ever moved,
and thus, there in no way that point Telescope, still/now at 11 in S has physically been placed at 4,0 in S.

Please do not give me the statement back that this is because you thought of moving to the Desk from your Bed, or that one frame has moved a distance of vt from the other. One frame has moved. No points within either frame have altered location within their own frame !! This is why I said no points have moved ( as just one frame, S', has moved )

What you miss, is that by virtue of allowing x' IN S...to be 11-7,
according to what x' = x- vt means, means in S, Einstein assigns a value of 4 to x' in S. I understand that you do not grasp this; ...why/how this equation assigns 4 to x' in S. I feel like I am saying 1 + 1 = 2, and you say, nope...1 + 1 = 1. We have reached a dead end dude.

Please, anyone else....got two cents to throw in ?...does x' = x - vt actually mean that P' in S equals 4 ?

This is where you and I differ.
Einstein's math says P' in S is 4...since x' = x - vt, and since 11-7 = 4.

Until you see/get/grasp/comprehend/understand/accept this,
then any further discussion with you is gonna be redundant. We will just keep repeating our stances and ultimately start posturing and writing in CAPS to add juice to our claims.

This is where we agree.
Einstein's math says P in S is 11
Einstein's math says P in S is P' in S.

Hmmmm, I wonder why not a single comment as of yet by Miles Mathis on this thread. He posted his pdf here at gsj on this very subject...twice. I believe he even wrote to JPL and they rejected him because he failed to deal with their concerns about motion.

I wrote him 4 weeks ago, and apparently he said he was too busy formulating his efforts for NASA to show that pi was equal to 4. He was too busy to give a response to when did he come up with his work on the Galilean. So not a single word to me about the Galilean ever.

I should be talking with Miles Mathis here...a believer. So I wait I guess, for pi to be proven to be equal to 4 first.

steve waterman

lol

I feel like I'm taking crazy pills.


The Galilean transformation is wrong, but not for those reasons.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Telescope is four from Desk in S'.
After transformation, the POINT telescope,
must be at four from the Bed in S.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
No. The transformation is from the Bed coordinates to the Desk. Not the telescope.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Since this has not happened...then point Telescope was never ever moved,
and thus, there in no way that point Telescope, still/now at 11 in S has physically been placed at 4,0 in S.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Exactly! The telescope is not at 4,0 in S, it is at 11 in S, that is what I've been saying all along!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Please do not give me the statement back that this is because you thought of moving to the Desk from your Bed, or that one frame has moved a distance of vt from the other. One frame has moved. No points within either frame have altered location within their own frame !! This is why I said no points have moved ( as just one frame, S', has moved )
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Exactly, the points don't move, only the frame moves.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


What you miss, is that by virtue of allowing x' IN S...to be 11-7,
according to what x' = x- vt means, means in S, Einstein assigns a value of 4 to x' in S. I understand that you do not grasp this; ...why/how this equation assigns 4 to x' in S. I feel like I am saying 1 + 1 = 2, and you say, nope...1 + 1 = 1. We have reached a dead end dude.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I don't allow x' to exist in S. Einstein didn't write the equation, Galileo or Newton did. x' only exists in reference to S'.

I'm not saying 1+1 = 1, please don't insult me when I've clearly got enough of an understanding of the math to continue arguing it purely on the terms laid out.

You're saying 1 + 1 = 2, so 1 = 2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



Please, anyone else....got two cents to throw in ?...does x' = x - vt actually mean that P' in S equals 4 ?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
No, it means that P' in S equals the x coordinate - vt. P' = 4 + 7 in S
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



This is where you and I differ.
Einstein's math says P' in S is 4...since x' = x - vt, and since 11-7 = 4.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
It wasn't Einstein's math. He didn't say that, he said P' in S is 4 + 7
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Until you see/get/grasp/comprehend/understand/accept this,
then any further discussion with you is gonna be redundant. We will just keep repeating our stances and ultimately start posturing and writing in CAPS to add juice to our claims.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I'm not going to agree that 11 = 4
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


This is where we agree.
Einstein's math says P in S is 11
Einstein's math says P in S is P' in S.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
It wasn't Einstein's math. P in S is 11, P in S is P' in S'.
P' in S is 4' + 7 = 11. S' in S is +7. P' in S' is 4'.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



Hmmmm, I wonder why not a single comment as of yet by Miles Mathis on this thread. He posted his pdf here at gsj on this very subject...twice. I believe he even wrote to JPL and they rejected him because he failed to deal with their concerns about motion.

I wrote him 4 weeks ago, and apparently he said he was too busy formulating his efforts for NASA to show that pi was equal to 4. He was too busy to give a response to when did he come up with his work on the Galilean. So not a single word to me about the Galilean ever.

I should be talking with Miles Mathis here...a believer. So I wait I guess, for pi to be proven to be equal to 4 first.

steve waterman
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Pi only equals 4 if you trace a circle on a highly curved manifold.

Write to Miles Mathis, and explain your idea to him. Maybe he'll correspond with you. He's the one who has figured out the true value of pi. He also wrote a book about wine tasting. Wonder if there is a light cone connecting the two?

Steve, "I should be talking with Miles Mathis here...a believer. So I wait I guess, for pi to be proven to be equal to 4 first."

I have always felt it was somewhat...let's say (strange) for lack of a better word, that pi should be used to find the area of a circle expressed in square "inches", I think that if pi is to be used it should find the area in square "pi's".

I'd go crazy talking to him, attempting to place an exact value like Pi for that is silly.

Max; certainly, he can use angular acceleration instead of the circular circumference to obtain π.
However, the logic itself of this procedure is clearly circular; since π is implicit from the beginning
in deriving the equation for angular acceleration in the first place.

"I think that if pi is to be used it should find the area in square "pi's".



Actually, shouldn't the metric be: "pi inches?"

I think I remember that pi is also the ratio of the perimeter of a square to the circumference of its inscribed circle. Thus, the square measured in inches and the circumference measured in inches will give pi inches times the diameter in inches delivers square inches, right?